Find range when graph crosses horizontal asymptote and vertex of parabola btwn vertical asymptote

[JenB]

Please help mom who hasn't done algebra or calculus in decades trying to help daughter. I could not find anything helpful covering this.

f(x) = (-3x^2+3x+36) / (x^2-4)

She determined Vertical Asymptotes at x=-2 and x=2 Horizontal Asymptote HA at y=-3 with x-intercepts at (-3,0) and (4,0) and y-intercept at (0,-9) as well as the graph crosses the HA at x=-8 and returns to approach y=-3 from below

She sketched graph using above determinants as: x>2 a negative increasing slope curve approaching HA -3 from the top, and
A negative parabola shape with vertex near y-intercept and extends down to -ve infinity between VAs -2<x<2 and
A curve with increasing slope from (-8,-3) as -8<x<2 and y approaches +ve infinity. But as this curve extends past (-8,-3) HA-intercept as x approaches -ve infinity y re-approaches the HA from below.

How do you define the range exactly without randomly guesstimating? i.e. For the parabola shape between -2<x<2 can vertex (x,A) be determined where range A>y>=infinity? And how far down does y <-3 HA when x<-8?

 

[Dr.Peterson]

Quick answer: calculus.

In a precalculus course, you would not be asked for the exact range; at most, you might be asked to use a graphing calculator to find the approximate range.

 

[JenB]

Hi Dr. Peterson,

Sorry for my confusion- I posted the question out of desperation last night because I'm trying to help my daughter if/when required to correct her tests. In this case the teacher never took up the questions so we didn't know what was expected. I'm now thinking also that my daughter failed to inform me that they were using desmos.com (something I just found) since she doesn't have a graphing calculator.

Anyways I just figured out that using the discriminant > 0 (because there are two x-intercepts) as a related inequality and the quadratic formula helps estimate the range if that's what the teacher expects.

Thank you for putting my heart at ease because I don't remember doing such questions the long way when I was 16.
J

 

[Dr.Peterson]

I'm not sure what discriminant you are referring to, but for some rational equations it is possible to use the discriminant of a quadratic obtained by inverting the function, so that may be what you are doing. (I just tried it and found that it gives the same result as Desmos -- but in radical form, so it is exact, not just just an estimate.)

I have no idea what way might be expected without seeing the exact wording of the problem, and any examples the book gives; but as I suggested, I have not seen a precalculus book that teaches this, or that demands it for such relatively unpleasant functions, even though as you found it is possible to do.

 

[JenB]

Exact wording on test: "Graph the following function f(x)= (-3x^2 + 3x + 36) / (x^2 - 4) on the grid provided showing all key features" and below "Domain ________ Range_________". The mistake that cost 2 out of 3 marks was my daughter defined range as {y€R, y#-3} where I typed # is substituting for 'cannot equal'. That's what set me off on the search for the two numbers needed to correctly answer the range.

I'm not sure if this is the same as inverting the function, but the solution I arrived at was to:
Use y for f(x) and re-wrote in standard form (y+3)x^2 - 3x - 4y - 36 = 0
Plugged the b, a and c from the above standard form into the discriminant " b^2 - 4ac" set to >0 because we have two x-intercepts.
Put that discriminant equation into the standard form 16y^2 + 192y + 437 > 0 and used the quadratic formula to find y values that satisfy the inequality.
Is this legitimate?

Can you please share how to get a more exact answer solution using the inverted function method? I don't have access to their textbook so I have no clue how and how well the textbook prepares them for the test questions. Since I'm making her correct her mistakes on the tests/quizzes I figure I'd better know what the correct answer is and how to get it.

 

[Dr.Peterson]

What you did is exactly what I did (though one of us made a small arithmetic error, as I had 441 for your 437). Solving the inequality using the quadratic formula is perfectly legitimate and gives exact values for the boundary of the range. It's the only method I can think of for exact answers before calculus; if I had access to the book I'd be looking for examples to confirm what they want.

If the result were nicer, I would be more likely to think this is what they expect students to do. But the problem clearly expects manual graphing, which would not yield very close approximations of the range without a lot of work. So maybe this is what they want.

 

[JenB]

Thank you so much Dr. Peterson, that helps me to feel a bit more confident that I can somewhat help her to reason (logic) concepts through without being dependent on only memorizing - at least that's my goal. It's been a very steep relearning curve for me!